Question

scores on the gre (graduate record examination) are normally distributed with a mean of 503 and a standard deviation of 130. use the 68 - 95 - 99.7 rule to find the percentage of people taking the test who score between 113 and 503.
the percentage of people taking the test who score between 113 and 503 is 49.85%

Explanation:

Step1: Calculate the number of standard - deviations.

First, find the difference between the mean ($\mu = 503$) and the given score ($x = 113$). The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $\sigma = 130$. So, $z=\frac{113 - 503}{130}=\frac{- 390}{130}=-3$.

Step2: Apply the 68 - 95 - 99.7 Rule.

The 68 - 95 - 99.7 Rule states that about 99.7% of the data lies within 3 standard deviations of the mean in a normal distribution. That is, $P(-3

Answer:

$49.85\%$